$\operatorname{gcd}(ab,a+b)=1$ if $a$ and $b$ are relatively prime
I'm trying to show that if $\operatorname{gcd}(a,b) = 1$, then $\operatorname{gcd}(ab,a+b)=1$.
I've tried to use the gcd properties: $$\operatorname{gcd}(a,b)=1 \implies \operatorname{gcd}(a,a+b)=1\\\operatorname{gcd}(a,b)\implies\operatorname{gcd}(ab,b^2)=b$$ but I got stuck. Any hint will help.
Solution 1:
Hint:
Supose $k:=\gcd(ab,a+b) > 1$. Let $p$ be a prime that divides $k$.
Then $p$ divides $ab$ wich means that it divides $a$ or $b$ (but not both because $\gcd(a,b) = 1$.
Can $p$ divide $a+b$?